$f\text{'}\left(x\right)=3x^2-4$ and $f\text{'}\left(1\right)=-1$

$g\text{'}\left(x\right)=4x^4+6x^2-10x+12$ and $g\text{'}\left(1\right)=12$

$s\text{'}\left(t\right)=60-\mathrm{9,8}t$ and $s\text{'}\left(1\right)=\mathrm{50,2}$

$H\text{'}\left(t\right)=4t$ and $H\text{'}\left(1\right)=4$

$\frac{\text{d}y}{\text{d}x}=-2x+6$ and $y\text{'}\left(1\right)=4$

$P\text{'}\left(x\right)=3a{x}^2+2bx+c$ and $P\text{'}\left(1\right)=3a+2b+c$

$\frac{\text{d}TW}{\text{d}q}=\mathrm{1,5}{q}^2-12q-25$ and $TW\text{'}\left(1\right)=\mathrm{35,5}$

$K\text{'}\left(x\right)=9a{x}^2-6x-3a^2$ and $K\text{'}\left(1\right)=9a-6-3a^2$

$f\text{'}\left(x\right)=2x^3-8x$ and $f\text{'}\left(x\right)=0$ as $x=0\vee x=\pm 2$, so $(0,0)$, $(-2,-8)$ and $(2,-8)$.

$TW\text{'}\left(q\right)=-3q^2+6q+3$ and $TW\text{'}\left(q\right)=0$ as $q=1\pm \sqrt{2}$, so $(\mathrm{2,4};\mathrm{16,7})$ and $(\mathrm{-0,4};\mathrm{5,3})$.

$v\text{'}\left(t\right)=3t^2-4t+1$ and $v\text{'}\left(t\right)=0$ as $t=\frac{1}{3}\vee t=1$, so $(\frac{1}{3},\frac{4}{27})$ and $(1,0)$.

$TW\text{'}\left(p\right)=40-\mathrm{0,04}p$ and $TW\text{'}\left(p\right)=0$ as $p=1000$, so $(1000,20000)$.

$f\left(x\right)=(x^2-4)(x^2-9)=0$ gives you $x^2=4\vee x^2=9$ and $x=\pm 2\vee x=\pm 3$, so $(\pm 2,0)$ en $(\pm 3,0)$.

$f\left(x\right)=x^4-13x^2+36$ and $f\text{'}\left(x\right)=4x^3-26x$.

Tangent for $x=-2$ is $y=20x+40$.
Tangent for $x=2$ is $y=-20x+40$.
Intersect $(0,40)$.

$f\text{'}\left(x\right)=0$ if $x=0\vee x=\pm \sqrt{\mathrm{8,5}}$.

It allows you to find the three extrema: max. $f\left(0\right)=36$, min. $f\left(\mathrm{-}\sqrt{\mathrm{8,5}}\right)=\mathrm{-6,25}$ and min. $f\left(\sqrt{\mathrm{8,5}}\right)=\mathrm{-6,25}$.

$h\left(0\right)=\mathrm{0,5}$ m.

$h\text{'}\left(x\right)=\mathrm{-0,02}x+\mathrm{0,2}$ and $h\text{'}\left(0\right)=\mathrm{0,2}$ m/s

It is the speed with which the height of the object changes at $x=0$. It is therefore not the initial speed of the object itself!

$h\text{'}\left(x\right)=0$ gives you $x=10$, which corresponds to the point $(10;\mathrm{1,5})$.

It is the speed with which the height of the trajectory changes that is $0$. But there is also a component of the speed in the forward direction.

$GTK\left(q\right)=\frac{1200}{q}+\mathrm{0,2}q$

$q=0$; when $q=0$ you cannot determine the average cost.

min. $ATC\left(77\right)\approx \mathrm{30,98}$

$ATC\to \mathrm{0,2}$ as $q\to \infty$( $\infty$ is the symbol for "infinitely large" ).
The production costs per unit are eventually changing with the (fixed) costs per article.